\(\int \frac {(1-2 x)^2 (3+5 x)^2}{(2+3 x)^7} \, dx\) [1268]

Optimal result

Integrand size = 22, antiderivative size = 56 \[ \int \frac {(1-2 x)^2 (3+5 x)^2}{(2+3 x)^7} \, dx=-\frac {49}{1458 (2+3 x)^6}+\frac {518}{1215 (2+3 x)^5}-\frac {503}{324 (2+3 x)^4}+\frac {740}{729 (2+3 x)^3}-\frac {50}{243 (2+3 x)^2} \]

[Out]

-49/1458/(2+3*x)^6+518/1215/(2+3*x)^5-503/324/(2+3*x)^4+740/729/(2+3*x)^3-50/243/(2+3*x)^2

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 56, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.045, Rules used = {90} \[ \int \frac {(1-2 x)^2 (3+5 x)^2}{(2+3 x)^7} \, dx=-\frac {50}{243 (3 x+2)^2}+\frac {740}{729 (3 x+2)^3}-\frac {503}{324 (3 x+2)^4}+\frac {518}{1215 (3 x+2)^5}-\frac {49}{1458 (3 x+2)^6} \]

[In]

Int[((1 - 2*x)^2*(3 + 5*x)^2)/(2 + 3*x)^7,x]

[Out]

-49/(1458*(2 + 3*x)^6) + 518/(1215*(2 + 3*x)^5) - 503/(324*(2 + 3*x)^4) + 740/(729*(2 + 3*x)^3) - 50/(243*(2 +
3*x)^2)

Rule 90

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Int[ExpandI
ntegrand[(a + b*x)^m*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, p}, x] && IntegersQ[m, n] &&
(IntegerQ[p] || (GtQ[m, 0] && GeQ[n, -1]))

Rubi steps \begin{align*} \text {integral}& = \int \left (\frac {49}{81 (2+3 x)^7}-\frac {518}{81 (2+3 x)^6}+\frac {503}{27 (2+3 x)^5}-\frac {740}{81 (2+3 x)^4}+\frac {100}{81 (2+3 x)^3}\right ) \, dx \\ & = -\frac {49}{1458 (2+3 x)^6}+\frac {518}{1215 (2+3 x)^5}-\frac {503}{324 (2+3 x)^4}+\frac {740}{729 (2+3 x)^3}-\frac {50}{243 (2+3 x)^2} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 31, normalized size of antiderivative = 0.55 \[ \int \frac {(1-2 x)^2 (3+5 x)^2}{(2+3 x)^7} \, dx=-\frac {8198+8172 x+52515 x^2+248400 x^3+243000 x^4}{14580 (2+3 x)^6} \]

[In]

Integrate[((1 - 2*x)^2*(3 + 5*x)^2)/(2 + 3*x)^7,x]

[Out]

-1/14580*(8198 + 8172*x + 52515*x^2 + 248400*x^3 + 243000*x^4)/(2 + 3*x)^6

Maple [A] (verified)

Time = 2.30 (sec) , antiderivative size = 29, normalized size of antiderivative = 0.52

method	result	size
norman	\(\frac {-\frac {50}{3} x^{4}-\frac {460}{27} x^{3}-\frac {389}{108} x^{2}-\frac {227}{405} x -\frac {4099}{7290}}{\left (2+3 x \right )^{6}}\)	\(29\)
gosper	\(-\frac {243000 x^{4}+248400 x^{3}+52515 x^{2}+8172 x +8198}{14580 \left (2+3 x \right )^{6}}\)	\(30\)
risch	\(\frac {-\frac {50}{3} x^{4}-\frac {460}{27} x^{3}-\frac {389}{108} x^{2}-\frac {227}{405} x -\frac {4099}{7290}}{\left (2+3 x \right )^{6}}\)	\(30\)
parallelrisch	\(\frac {12297 x^{6}+49188 x^{5}+49980 x^{4}+40160 x^{3}+29520 x^{2}+8640 x}{1920 \left (2+3 x \right )^{6}}\)	\(39\)
default	\(-\frac {49}{1458 \left (2+3 x \right )^{6}}+\frac {518}{1215 \left (2+3 x \right )^{5}}-\frac {503}{324 \left (2+3 x \right )^{4}}+\frac {740}{729 \left (2+3 x \right )^{3}}-\frac {50}{243 \left (2+3 x \right )^{2}}\)	\(47\)
meijerg	\(\frac {3 x \left (\frac {243}{32} x^{5}+\frac {243}{8} x^{4}+\frac {405}{8} x^{3}+45 x^{2}+\frac {45}{2} x +6\right )}{256 \left (1+\frac {3 x}{2}\right )^{6}}-\frac {x^{2} \left (\frac {81}{16} x^{4}+\frac {81}{4} x^{3}+\frac {135}{4} x^{2}+30 x +15\right )}{640 \left (1+\frac {3 x}{2}\right )^{6}}-\frac {59 x^{3} \left (\frac {27}{8} x^{3}+\frac {27}{2} x^{2}+\frac {45}{2} x +20\right )}{7680 \left (1+\frac {3 x}{2}\right )^{6}}+\frac {x^{4} \left (\frac {9}{4} x^{2}+9 x +15\right )}{384 \left (1+\frac {3 x}{2}\right )^{6}}+\frac {5 x^{5} \left (\frac {3 x}{2}+6\right )}{192 \left (1+\frac {3 x}{2}\right )^{6}}\)	\(135\)

[In]

int((1-2*x)^2*(3+5*x)^2/(2+3*x)^7,x,method=_RETURNVERBOSE)

[Out]

(-50/3*x^4-460/27*x^3-389/108*x^2-227/405*x-4099/7290)/(2+3*x)^6

Fricas [A] (verification not implemented)

none

Time = 0.22 (sec) , antiderivative size = 54, normalized size of antiderivative = 0.96 \[ \int \frac {(1-2 x)^2 (3+5 x)^2}{(2+3 x)^7} \, dx=-\frac {243000 \, x^{4} + 248400 \, x^{3} + 52515 \, x^{2} + 8172 \, x + 8198}{14580 \, {\left (729 \, x^{6} + 2916 \, x^{5} + 4860 \, x^{4} + 4320 \, x^{3} + 2160 \, x^{2} + 576 \, x + 64\right )}} \]

[In]

integrate((1-2*x)^2*(3+5*x)^2/(2+3*x)^7,x, algorithm="fricas")

[Out]

-1/14580*(243000*x^4 + 248400*x^3 + 52515*x^2 + 8172*x + 8198)/(729*x^6 + 2916*x^5 + 4860*x^4 + 4320*x^3 + 216
0*x^2 + 576*x + 64)

Sympy [A] (verification not implemented)

Time = 0.07 (sec) , antiderivative size = 51, normalized size of antiderivative = 0.91 \[ \int \frac {(1-2 x)^2 (3+5 x)^2}{(2+3 x)^7} \, dx=\frac {- 243000 x^{4} - 248400 x^{3} - 52515 x^{2} - 8172 x - 8198}{10628820 x^{6} + 42515280 x^{5} + 70858800 x^{4} + 62985600 x^{3} + 31492800 x^{2} + 8398080 x + 933120} \]

[In]

integrate((1-2*x)**2*(3+5*x)**2/(2+3*x)**7,x)

[Out]

(-243000*x**4 - 248400*x**3 - 52515*x**2 - 8172*x - 8198)/(10628820*x**6 + 42515280*x**5 + 70858800*x**4 + 629
85600*x**3 + 31492800*x**2 + 8398080*x + 933120)

Maxima [A] (verification not implemented)

none

Time = 0.23 (sec) , antiderivative size = 54, normalized size of antiderivative = 0.96 \[ \int \frac {(1-2 x)^2 (3+5 x)^2}{(2+3 x)^7} \, dx=-\frac {243000 \, x^{4} + 248400 \, x^{3} + 52515 \, x^{2} + 8172 \, x + 8198}{14580 \, {\left (729 \, x^{6} + 2916 \, x^{5} + 4860 \, x^{4} + 4320 \, x^{3} + 2160 \, x^{2} + 576 \, x + 64\right )}} \]

[In]

integrate((1-2*x)^2*(3+5*x)^2/(2+3*x)^7,x, algorithm="maxima")

[Out]

-1/14580*(243000*x^4 + 248400*x^3 + 52515*x^2 + 8172*x + 8198)/(729*x^6 + 2916*x^5 + 4860*x^4 + 4320*x^3 + 216
0*x^2 + 576*x + 64)

Giac [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 29, normalized size of antiderivative = 0.52 \[ \int \frac {(1-2 x)^2 (3+5 x)^2}{(2+3 x)^7} \, dx=-\frac {243000 \, x^{4} + 248400 \, x^{3} + 52515 \, x^{2} + 8172 \, x + 8198}{14580 \, {\left (3 \, x + 2\right )}^{6}} \]

[In]

integrate((1-2*x)^2*(3+5*x)^2/(2+3*x)^7,x, algorithm="giac")

[Out]

-1/14580*(243000*x^4 + 248400*x^3 + 52515*x^2 + 8172*x + 8198)/(3*x + 2)^6

Mupad [B] (verification not implemented)

Time = 0.03 (sec) , antiderivative size = 46, normalized size of antiderivative = 0.82 \[ \int \frac {(1-2 x)^2 (3+5 x)^2}{(2+3 x)^7} \, dx=\frac {740}{729\,{\left (3\,x+2\right )}^3}-\frac {50}{243\,{\left (3\,x+2\right )}^2}-\frac {503}{324\,{\left (3\,x+2\right )}^4}+\frac {518}{1215\,{\left (3\,x+2\right )}^5}-\frac {49}{1458\,{\left (3\,x+2\right )}^6} \]

[In]

int(((2*x - 1)^2*(5*x + 3)^2)/(3*x + 2)^7,x)

[Out]

740/(729*(3*x + 2)^3) - 50/(243*(3*x + 2)^2) - 503/(324*(3*x + 2)^4) + 518/(1215*(3*x + 2)^5) - 49/(1458*(3*x
+ 2)^6)